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Question

If  $$f(x)$$ =$$\displaystyle \frac{sin{3x} + A sin{2x} + B sinx}{x^5}   \,\,\,\,\, (x \neq 0)$$ is continuous at $$x = 0$$, then find $$A$$ +$$B$$.


Solution

$$\displaystyle f(x) = \underset{x \rightarrow 0}{lim} \frac{sin{3x} + Asin{2x} + B sin{x}}{x^5}$$

$$\displaystyle= \underset{x \rightarrow 0}{lim} \frac{sin{x}}{x} \left(\frac{3 - 4 sin^2{x} + 2Acos{x} + B}{x^4} \right) $$

$$\displaystyle= \underset{x \rightarrow 0}{lim} \frac{1 + 2cos{2x} + 2A cos{x} + B}{x^4}$$

$$\displaystyle= \underset{x \rightarrow 0}{lim} \frac{1}{x^4} \left( 1 + 2\left( 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} + ..... \right) + 2A \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \right) + B \right)$$

$$\displaystyle= \underset{x \rightarrow 0}{lim} \frac{1}{x^4} (3 + 2A + B + x^2(-4-A) + x^4 \left( \frac{4}{3} + \frac{A}{12} \right) + .....)$$

$$\Rightarrow$$ $$2A + B + 3 = 0$$ and $$-4 - A = 0$$
$$\Rightarrow$$ $$A = -4, B = 5$$

$$A+B = -4+5 =1$$

Mathematics

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